A Characterization of Bi-Jordan Homomorphisms on Banach Algebras

Abstract

For Banach algebras and , we show that if is unital and commutative, each bi-Jordan homomorphism from into a semisimple commutative Banach algebra is a bihomomorphism.

1. Introduction

Let and be complex Banach algebras and let be a linear map. Then is called -homomorphism if for all , The concept of -homomorphism was studied for complex algebras by Hejazian et al. in [1]. A 2-homomorphism is then just a homomorphism, in the usual sense. We refer the reader to [2], for certain properties of 3-homomorphisms.

In [3], Eshaghi Gordji introduced the concept of an -Jordan homomorphism. A linear map between Banach algebras and is called an -Jordan homomorphism if A 2-Jordan homomorphism is called simply a Jordan homomorphism.

It is obvious that each -homomorphism is an -Jordan homomorphism, but in general the converse is false. The converse statement may be true under certain conditions. For example, Zelazko in [4] proved that every Jordan homomorphism from Banach algebra into a semisimple commutative Banach algebra is a homomorphism. See also [5] for another approach to the same result. The reader is referred to [6], for characterization of 3-Jordan homomorphism.

Also it is shown in [3] that every -Jordan homomorphism between two commutative Banach algebras is an -homomorphism for and this result is extended to the case in [7]. Some investigation has been done about Jordan derivations and Jordan centralizers in [8, 9].

Throughout the paper, let . Then is a Banach algebra for the multiplication and with norm Clearly, is commutative if and only if both and are commutative, and it is unital if and only if both and are unital. Without any confusion we denote by , the unit element of both and .

Let be a complex Banach algebra. A bilinear map is a function such that for any the map is linear map from to , and for any the map is linear map from to .

A bilinear map is called bihomomorphism if for all , and it is called bi-Jordan homomorphism (BJH, for short) if

It is obvious that each bihomomorphism is BJH, but in general the converse is not true. For example, let and , where Define by . Then is bilinear and for all . Thus, is a bi-Jordan homomorphism, but it is not bihomomorphism. For instance, let Then and

In this paper, we show that the converse statement holds under certain conditions.

2. Main Results

We commence with some useful lemmas.

Lemma 1. Suppose that is a BJH. Then (1);(2).

Proof. The proof is straightforward.

Lemma 2. Suppose that is a BJH. If is unital and commutative, then (1);(2).

Proof. The proof follows from Lemma 1.

Lemma 3. Let be unital, and let be a nonzero BJH. Then .

Proof. Let be a BJH. Then for all , we get Replacing by and by in (11) gives Assume that ; then by Lemma 1, It follows from (12) and (13) that for all . By Lemma 1, Thus, by (14) and (15), we get , for all , which is contradiction.

Now we state and prove the main theorem. The main idea of the proof can be found in [6].

Theorem 4. Let be unital and commutative, and let be a BJH from into a semisimple commutative Banach algebra . Then is a bihomomorphism.

Proof. We first assume that and let be a BJH. Then, for all , we get Replacing by and by in (16), we have By Lemma 3, , so (16) gives It follows from Lemma 1 that Thus, by (17), (18), and (19) we get for all . Replacing by and by in (20) gives By (20) and Lemma 2 we deduce By (21) and (22), for all , and so is bihomomorphism.
Now suppose that is semisimple and commutative. Let be the maximal ideal space of . We associate with each a function defined by Pick arbitrary. It is easy to see that is a BJH, so by the above argument it is a bihomomorphism. Thus by the definition of we have Since was arbitrary and is assumed to be semisimple, we obtain for all .

The second dual space of a Banach algebra admits two Banach algebra multiplications known as the first and second Arens products, each extending the product on . These products which we denote by and , respectively, can be defined as follows: where and are nets in that converge, in -topologies, to and , respectively. The Banach algebra is said to be Arens regular if on the whole of . Some significant results related to the Arens regularity of certain bilinear maps and Banach algebra obtained in [10]. For more information on the Arens products, we refer the reader to [11, 12], for example.

It is shown in [11] that every -algebra is Arens regular and semisimple. Also the second dual of each -algebra is also a -algebra.

Theorem 5. If is the natural embedding of a Banach space into , then is -dense in .

Proof. See [13].

Theorem 6. Let be a continuous BJH. Then the second adjoint of is also a BJH.

Proof. By Theorem 5, there are bounded nets and in and that converge, in -topologies, to and , respectively. Then Thus, is BJH, as claimed.

Since the second dual of every -algebra is unital [14], we deduce the following result from Theorems 4 and 6.

Corollary 7. Let be a continuous BJH between commutative -algebras. Then is a bihomomorphism.

Theorem 8. Suppose that is a BJH from the unital Banach algebra into . Then is a -BJH; that is, for all .

Proof. Let be a BJH. Then, for all , Replacing by and by in (30), we have Replacing by and by in (31) gives Replacing by and by in (32), we obtain Replacing and by in (31), we get Since , so (34) gives for all . By Lemma 1, It follows from (35) and (36) that By (33) and (37), for all . Thus, the result is established for . An easy induction argument now finishes the proof.

As a consequence of Theorem 8 we have the next result.

Corollary 9. Suppose that is a BJH from the unital Banach algebra into a semisimple commutative Banach algebra . Then is an -BJH.

Baker in [15] proved that an almost multiplicative function is either bounded or multiplicative. Now we prove an analogous result of Baker’s theorem for bihomomorphism.

Theorem 10. Let and satisfy for all . Then either is bihomomorphism or for all .

Proof. Let and . Suppose that there exist such that , so that , where . Then we have hence Now make the induction assumption: Let Then whence and (43) is established for all .
Given that , we have Thus, Hence Therefore Now put to obtain for all and . Letting , we get and so is a bihomomorphism.

As Baker pointed out in his article, the above proof works for functions , where is a Banach algebra in which the norm is multiplicative; that is, for all . For norm algebra for which the norm is not multiplicative, the situation is false. For example, let and choose so that , and define by Then with the usual matrix norm for all . Clearly, is unbounded, but is not bihomomorphism.

Competing Interests

The author declares that there is no conflict of interests regarding the publication of this paper.